The function swap just swaps it's local copy of pointers which does not mean that it swap array elements. You have to do it explicitly.
On Fri, Jun 17, 2011 at 3:17 AM, udit sharma <[email protected]>wrote: > Ohh Sry... The qus was: > > >> #include<stdio.h>void swap(char *,char *);int main(){char *ps[2]={ >>> "Hello", >>> "Good Mornning", >>> };swap(ps[0],ps[1]);printf("%s \t %s\n",ps[0],ps[1]);return 0;} >>> void swap(char *p,char *q){char *t;t=p;p=q;q=t;} >>> >>> >>> why the output is: >>> Hello Good Mornning >>> >>> >>> >>> >>> >>> >>> Regards > UDIT > DU- MCA > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > Regards -- Rohit Sindhu -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
